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_Mayday_

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Hey, I would like to check that my working for the following question is along the right lines. I have an exam this coming Wednesday and there are a few really common questions that I think I should do some work on. As always any help is much appreciated!

A parcel of weight 10N lies on a rough plane inclined at an angle of 30 degrees to the horizontal. A horizontal force of magnitude

The horizontal force is removed.

[itex]18 - 10\cos30 - P\sin 30[/itex]

[itex]18 = 10\cos30 + P\sin 30[/itex]

[itex]18 - 5\sqrt3 = \frac{P}{2}[/itex]

[itex]2(18 - 5\sqrt3) = P = 18.7N[/itex]

[itex]F + 10\sin30 = 18.7\cos30[/itex]

[itex]5F = 16.20[/itex]

[itex]F = 3.24[/itex]

Now I just plug in the answer from

[itex]F=\mu R[/itex]

[itex]3.24 = 18\mu[/itex]

[itex]\mu = 0.18[/itex]

I am still working on

Thank you.

**The Question**A parcel of weight 10N lies on a rough plane inclined at an angle of 30 degrees to the horizontal. A horizontal force of magnitude

**P**Newtons acts on the parcel as shown in the attachment (Bottom of post). The parcel is in equilibrium and on the point of slipping up the plane. The normal reaction of the plane on the parcel is 18N. The coefficient of friction between the parcel and the plane is [itex]\mu[/itex]. Find:**(a)**The value of P**(b)**The value of [itex]\mu[/itex]The horizontal force is removed.

**(c)**Determine whether or not the parcel moves.**My Attempt****(a)**Resolve perpendicular to the plane:[itex]18 - 10\cos30 - P\sin 30[/itex]

[itex]18 = 10\cos30 + P\sin 30[/itex]

[itex]18 - 5\sqrt3 = \frac{P}{2}[/itex]

[itex]2(18 - 5\sqrt3) = P = 18.7N[/itex]

**(b)**Resolve along the plane.[itex]F + 10\sin30 = 18.7\cos30[/itex]

[itex]5F = 16.20[/itex]

[itex]F = 3.24[/itex]

Now I just plug in the answer from

**(a)**with this one from**(b)**[itex]F=\mu R[/itex]

[itex]3.24 = 18\mu[/itex]

[itex]\mu = 0.18[/itex]

I am still working on

**(c)**but could someone please have a look at how I am doing so far. I may have skipped a few steps, but I am just using the fact that when resolving perpendicular you can always use cos, and whenever you resolve parallel you can use sin, when dealing with vectors int his course on an inclined plane.Thank you.

**_Mayday_**